h=-16t2+80t+96
Use this position polynomial to calculate the following:
1.The height of the object after 2 seconds
2.The height of the object after 4 seconds
3.How long the object will take to reach the ground?
4.Write the position function for an object that is thrown upward from a 180 ft. building with an initial velocity of 20 ft/sec.
1. 192
2. 150
I am not sure how to do numbers 3 and 4 and I am not sure If I did 1 and 2 right?
Thanks in advance!
#1. ok
#2. I get 160 -- better check again
#3. solve for t when h=0
#4. 180 + 20t - 16t^2
To calculate the height of the object after a certain amount of time, we can substitute the given time values into the position polynomial equation h = -16t^2 + 80t + 96. Let's calculate the height for 2 seconds and 4 seconds:
1. For 2 seconds:
Substitute t = 2 into the equation: h = -16(2)^2 + 80(2) + 96
Simplifying the equation, we get: h = -64 + 160 + 96
h = 192 feet
2. For 4 seconds:
Substitute t = 4 into the equation: h = -16(4)^2 + 80(4) + 96
Simplifying the equation, we get: h = -256 + 320 + 96
h = 160 feet
So, you calculated the heights correctly for 2 seconds and 4 seconds as 192 feet and 160 feet, respectively.
Now, moving on to the remaining questions:
3. To find out how long it will take for the object to reach the ground, we need to determine when the height (h) is equal to zero. This represents the time at which the object hits the ground.
Setting h = 0 in the position polynomial: 0 = -16t^2 + 80t + 96
To solve this quadratic equation, you can either factor it or use the quadratic formula. Let's use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation a = -16, b = 80, and c = 96. Substituting these values into the quadratic formula:
t = (-(80) ± √((80)^2 - 4(-16)(96))) / (2 (-16))
Simplifying further, we get:
t = (-80 ± √(6400 + 6144)) / (-32)
t = (-80 ± √(12544)) / (-32)
t = (-80 ± 112) / (-32)
Using both values, we get t ≈ -4 or t ≈ 6.
Since we are interested in the positive time (the negative value does not make sense here), the object will take approximately 6 seconds to reach the ground.
4. To write the position function for an object thrown upward from a 180 ft. building with an initial velocity of 20 ft/sec, we can use a similar equation h = -16t^2 + vt + s, where h is the height, t is the time, v is the initial velocity, and s is the initial height or position.
Since the object is thrown upward, the initial velocity v will be negative (-20 ft/sec) because it is acting against gravity. Also, the initial position s will be 180 ft. Substituting these values into the equation, we get:
h = -16t^2 - 20t + 180
So, the position function for this scenario would be h = -16t^2 - 20t + 180.
I hope this helps! Let me know if you have any further questions.